Practice problems--1999
| Week 1 | Week 2 | Week 3 | Week 4 | Week 5 |
| Week 6 | Week 7 | Week 8 | Week 9 | Week 10 |
| Week 11 |
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About these problems Each week's problems run the scale from very simple to somewhat complex. The intent is to get you used to problem-solving. Questions on the homeworks and exams will be on the complex end of the scale. For samples of homework/exam problems, look in the back of the course notes package and here. |
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| Week 11 |
Lecture 35
| 1. | A colony of black cats produces a total of 400 kittens one year. One of the kittens turns out to have a dominant mutation, giving the kitten baby blue polka dots. What is the probability that the wild type allele will be lost due to genetic drift in the next generation? | |
| 2. | A different colony of the same species of cats has 400 members, 396 of whom show the polka-dot phenotype. This colony merges with a larger colony in which 546 out of 600 members are polka dotted. What will be the frequency of black cats in the next generation? |
| 3. | In yet another colony of these infernal cats, the frequency of plain black cats is 0.25. | |
| (i) | If the colony is infected by a lethal virus that only kills polka-dotted individuals, what will be the frequency of the black allele in the next generation? | |
| (ii) | If the virus only kills plain black cats, what will be the frequency of black cats in the next generation? (Assume that the virus has disappeared by then.) | |
| 4. |
From 1998 A population is entirely homozygous for allele D. However, D can mutate to d at a rate of 0.00012 per generation, while d mutates to D at a rate of 0.00004 per generation. The population is otherwise at Hardy-Weinberg conditions. |
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| (i) | Given infinite time, which allele do you think will eventually predominate in the population? Why? | |
| (ii) | What is the change in the frequency of D per generation? How about the change in the frequency of d? (Remember that for each part, you have to take into consideration both the forward mutation and the back mutation.) | |
| (iii) | What will the two allele frequencies be when they reach an equilibrium (increase in frequency of d by forward mutation matched by decrease due to back mutation)? | |
| 5. | For you to think about... | |
| (i) | In a population of 1000 individuals, 250 members are homozygous recessive bb. What is the predicted number of BB and Bb individuals if this population shows Hardy-Weinberg frequencies? [That's the easy part.] | |
| (ii) |
And now the real question -- The actual number of BB individuals was found to be 220, and the number of Bb individuals was 580. If you wanted to do a chi-square test to see if these numbers are consistent with Hardy-Weinberg predictions, what value would you use for df (# of degrees of freedom)? Why? |
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Lectures 36 and 37
| 6. | One of the problems using DNA tests to screen for carriers of cystic fibrosis is that the test has only ~70% success rate in detecting carriers (because there are many different mutant alleles in the population). For the following questions, assume that the frequency of carriers = 0.05. | |
| (i) | If all potential marriage partners are screened for carrier status, what fraction of heterozygote/heterozygote pairs would be missed? (i.e., not recognized as heterozygote x heterozygote pairs). | |
| (ii) | If only one member in a couple is tested, and is found to be a heterozygote, what is the probability that both members of the couple are heterozygotes? | |
| (iii) | If one member is found to be a heterozygote, and the other member is also tested and found not to have disease alleles, what is the probability that both members of the couple are heterozygotes? | |
| 7. | A young boy has acute familial hypercholesterolemia because both copies of the LDL receptor gene have promoter mutations that block their transcription. The family physician is well aware of the use of bile acid-binding resins in treating hypercholesterolemia, but she tells the family that this treatment would be ineffective in this case. What was her (correct) reasoning? |
| 8. |
As a last resort in treating a dominant disease, some clinical researchers decide to use an antisense construct to block the disease gene function. The mutant allele (asterisk marks the mutation) and two possible antisense constructs (Construct 1 and Construct 2) are shown. Which of these two constructs has a better chance succeeding as an antisense construct, and why?
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| Week 10 |
Do #3, #8, and #13 for Quiz Section
Lecture 30
| 1. | In Drosophila, torso and fs are both maternal effect genes with the phenotype that homozygous mutant females produce embryos that lack tail structures. If you were given a strain of Drosophila that showed this phenotype (i.e., females produced tailless offspring), how would you determine which gene was mutated? |
| 2. | What change in the expression pattern of Krüppel and knirps would you expect for a mutant that has elevated levels of both bicoid and nanos? (Assume that the increased transcription of hunchback is exactly matched by increased inhibition of hunchback translation.) |
| 3. | The second thoracic segment in Drosophila is supposed to produce wings, while the third thoracic segment is supposed to produce halteres (flight balancers). A new homeotic mutation mut1 causes wings to develop on both second and third thoracic segments (no halteres anywhere), while a second mutation mut2 causes halteres to develop on both segments (no wings anywhere). Based on what you know about how segment identity is set, state whether you expect each mutant phenotype to be dominant or recessive, and why. |
Lecture 31
| 4. | For each of the following traits, state which is more important in determining the phenotype--genetic variation or environmental factors: | |||||||||||||||
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| 5. | In the city of Metropolis, all schools are equally good, children are impartially encouraged in mental tasks, and all the other environmental factors relevant to IQ tests are constantly good; i.e., there is no variation in environment within the city of Metropolis with regard to school education. In Gotham City (in the same country), all schools are equally bad, learning is never encouraged, and all the environmental factors relevant to taking IQ tests are constantly unfavorable; i.e., there is no variation in environment with respect to education within Gotham City. | |
| (i) | What is the percent heritability of IQ within Metropolis and within Gotham City? Explain. | |
| (ii) | What could be the cause of differences in IQ scores when comparing between Metropolis and Gotham city? | |
Lecture 32
Questions 6-9 are reproduced from the lecture notes (page preceding p.133).
| 6. | Assume that height in a plant is controlled by two gene pairs and that each additive allele contributes 5 cm to a base height of 20 cm (i.e., aabb = 20 cm). | |
| (i) | What is the height of an AABB plant? | |
| (ii) | Predict the phenotypic ratios of F1 and F2 plants in a cross between aabb and AABB. | |
| (iii) | List all the genotypes that give rise to plants that are 25 and 35 cm in height. | |
| 7. | In a cross where three gene pairs determine weight in squash, what proportion of individuals from the cross AaBbCC x AABbcc will contain only 2 additive alleles? What genotype(s) fall into this category? |
| 8. | An inbred strain of plants has a mean height of 24 cm. A second strain of the same species from a different country also has a mean height of 24 cm. The F1 plants from a cross between these two strains are also 24 cm high. However, the F2 generation shows a wide range of heights; the majority are like the P1 and F1 plants, but approximately 4 of 1000 are only 12 cm high, and 4 of 1000 are 36 cm high. | |
| (i) | What mode of inheritance is occurring here? | |
| (ii) | How many gene pairs are involved? | |
| (iii) | How much does each gene contribute to the plant height? | |
| (iv) | Indicate one possible set of genotypes of the P1 and F1 plants that could explain their heights. | |
| (v) | Indicate one possible set of genotypes to account for F2 plants that are 18 cm or 33 cm high. | |
| 9. | Plants may be 10, 20, 30, 40, 50, 60 or 70 cm high where plant height is under polygenic control. A 10-cm true-breeding plant is crossed to a 50-cm true-breeder. How many gene pairs are involved? What F1 and F2 results can be predicted? |
Lecture 33
| 10. | In a certain population of 20 million people, 500 individuals are found to have disaccharide intolerance, an autosomal recessive trait. Assuming that this number reflects Hardy-Weinberg frequencies, how many individuals in the population are expected to be carriers of the trait? |
| 11. | Island iguanas come in two varieties, beach-loving (dominant phenotype) and bridge-loving (recessive phenotype). The frequency of bridge-loving iguanas on one island is 0.04. On a neighboring island that has an equal-sized population of iguanas, the frequency of bridge-loving iguanas is 0.16. One day some bridge-loving iguanas build a bridge between the two islands, so now the iguanas can move freely between islands. | |
| (i) | Suppose the iguanas on each island mate exclusively with iguanas from the other island, what will be the frequency of bridge-loving iguanas in the next generation? How about the generation after that? | |
| (ii) | Suppose, instead, that the beach-loving iguanas on one island mated only with beach-loving iguanas on the other island , and that bridge-loving iguanas on one island mated only with bridge-loving iguanas on the other island, what would be the frequency of beach-loving and bridge-loving iguanas in the next generation? | |
| 12. | What is the relationship between genotype frequency and allele frequency in females vs. males for an X-linked recessive trait? |
| 13. | Pattern baldness is an autosomal trait in humans that shows sex-influenced inheritance -- it is dominant in males (BbBh and BbBb males become bald) but recessive in females (only BbBb females become bald). In a certain population that shows Hardy-Weinberg frequencies, 9% of the women become bald. | |
| (i) | What percentage of the men become bald? | |
| (ii) | What will the frequency of allele Bb among men in the next generation? | |
| Questions from yesteryear | |||||||||||||||||
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| Week 9 |
Do #2, #4, and #6 for quiz section.
Lectures 27, 28
| 1. | What phenotypes would you predict would result from each of the following mutations in E. coli? Assume in each case that the cell is otherwise wild type. | |
| (a) | lacP- (promoter deleted) | |
| (b) | lacOc | |
| (c) | lacA- (lacA deleted) | |
| (d) | lacY- (missense mutation) | |
| (e) | lacY- (STOP codon near the start of the gene) | |
| (f) | CAP- (CAP gene deleted) | |
| (g) | eno- (ENO is required for synthesis of phosphoenol pyruvate) | |
| 2. | For each partially diploid strain shown below, indicate whether beta-galactosidase activity will be inducible (normal induction), constitutively low (i.e., low in the presence or absence of inducer), or constitutively high (high in the presence or absence of inducer). Give a brief explanation for each answer. Assume that glucose is never present. | |
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| (ii) |
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| (iii) |
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| (iv) |
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| (v) |
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| As discussed in class, i- mutations prevent binding of repressor to the operator; is mutations result in "super-repressors" that cannot be dislodged from the operator; p- mutations prevent transcription initiation from the promoter; oc mutations prevent binding of repressor to the operator, and z- mutations cause production of non-functional beta-gal. | ||
| 3. | You heard in lecture that galactose can inactivate the Gal80 protein, allowing transcription of galactose-inducible genes. In fact, Gal80 does not respond directly to galactose; rather, in response to galactose, Gal3 protein binds and inactivates Gal80 protein. State whether you expect each of the following GAL3 mutations to give a dominant or recessive phenotype: | |
| (a) | gal3c : binds and inactivates Gal80 whether galactose is present or absent | |
| (b) | gal3- : incapable of binding to Gal80 protein | |
| 4. |
The diagram below represents an inducible operon in a certain bacterium; A, B, and C represent three structural genes. Control elements consist of promoter, operator, etc.
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| (a) | A mutation in gene B results in lack of functional proteins B and C, but protein A is functional. Explain this phenotype. | |
| (b) | Regulatory mutations were found in a separate gene, reg. The reg-1 mutation results in constitutively high expression of operon ABC; the reg-2 mutation results in constitutively low expression of ABC. The use of partially diploid bacteria reveals that the reg-1 phenotype is dominant, while the reg-2 phenotype is recessive. Propose a mechanism for regulation of operon ABC. | |
Lecture 29
| 5. | A certain mutation in the toad Xenopus is lethal. Examination of developing embryos shows that early development is normal and indistinguishable from wildtype, but development stalls after the first few cell divisions. Based on what you know about development in Drosophila, suggest a reason for the delayed effect of the Xenopus mutation. |
| 6. | Predict the phenotypes of the following homozygous null mutations in Drosophila. | |
| (a) | nanos- | |
| (b) | hunchback- (assume here that that this is a purely zygotic gene) | |
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Also try: 1998 Problem set 5, Q. 6 (in back of lecture notes) |
| Week 8 |
Do #3, #5, and #12 for quiz section
Lecture 24
| 1. | Two true-breeding, blind strains of crickets were crossed to each other. The F1 crickets all showed normal sight. When these F1 crickets were crossed to each other, the resulting F2 progeny consisted of 178 crickets with normal vision and 142 blind crickets. | |
| (i) | Are the two original parental strains mutated in the same gene? If not, what is the mimimum number of genes involved? | |
| (ii) | Explain the results (why the F1 crickets could see, and why this ratio of F2 phenotypes was obtained). What progeny ratio would you expect if the F1 crickets were crossed to fully homozygous recessive crickets? | |
| (iii) | What fraction of the F2 would you expect to be true-breeding? | |
| 2. | In a certain breed of dogs, B and b determine black (dominant) and brown (recessive) coat color, respectively. The recessive allele e of a separate (independently assorting) gene blocks expression of both B and b alleles, giving a yellow coat. (The dominant E allele does not affect coat color.) Determine the parental genotypes (giving reasons) for each of these crosses: | |
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| 3. |
A new species of Great Northwestern Pygmy Rats was discovered in some caves near Ellensburg. Analysis of blood samples showed that like humans, these Ellensburg rats came in A, B, AB, or O blood types; as in humans, blood type is determined by addition of A-type or B-type sugars to a protein on the surface of blood cells. When scientists crossed a true-breeding B bloodtype rat with a true-breeding O rat, however, they were surprised to find that the progeny all had the AB blood type. The F2 progeny obtained by crossing these F1 AB rats to each other had the following blood groups:
Explain the results, using Punnett squares to illustrate your answer. Hint: Do the results resemble a monohybrid ratio or a dihybrid ratio? |
Lecture 25
| 4. | As you've heard previously, yeast cells that have a normal ADE gene (i.e., Ade+ cells) can grow in the absence of adenine in the medium, and produce white colonies on agar plates. Cells that are ade- require adenine in the medium, and produce red colonies on plates. Given a yeast strain that has a frameshift mutation in the ADE gene (rendering the gene non-functional), outline a selection and a screen to identify revertants that have a functional ADE gene. |
| 5. | A lab in Seattle is interested in understanding the genetics of flower petal development in petunias. After doing mutagenesis on the plant and screening the mutant products, they identify three homozygous recessive mutant strains that all fail to form petals; they call these mutants strains p1 through p3. Meanwhile, they hear from their friends in Sedro Woolley, who had done a similar screen and identified four recessive mutant alleles r1 through r4 that also failed to form flower petals. The labs exchange strains and perform crosses between the various strains; the ability (+) or inability (-) of the progeny to form petals is indicated: | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
How many genes are represented in this collection of mutant strains? Which mutants represent alleles of the same genes? Why is half the table above left blank? |
Lecture 26
| 6. |
The following is the pathway for synthesis of E, an essential metabolite of yeast:
Which compound or compounds (amongst A - E) will allow growth of yeast lacking-- |
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| (i) | enzyme E3? | |
| (ii) | enzyme E4? | |
| (iii) | enzymes E2 and E3? | |
| Which compound do you think would accumulate in each of the above mutants? | ||
| 7. |
The following branched pathway depicts the synthesis of E and F, two essential amino acids in a certain mold.
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| (i) | Name two compounds (amongst A-F) that will allow growth of a mutant lacking enzyme E3. | |
| (ii) | Name one compund that will allow growth of a mutant lacking enzyme E1. | |
| 8. |
Purple flower color in a plant species requires the conversion of a white precursor to red pigment by enzyme E1 and to blue pigment by enzyme E2. The combination of the two pigments gives purple color, as indicated:
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| (i) | What will be the phenotype of a plant homozygous for a null allele of the gene for E1? | |
| (ii) | What will be the phenotype of a plant homozygous for a null allele of the gene for E2? | |
| (iii) | What will be the F1 phenotype(s) if the plant in (i) is crossed to the plant in (ii)? | |
| (iv) | What will be the F2 phenoptypes and ratio if the F1 plants are crossed to each other? | |
| 9. |
Yeast that were capable of synthesizing the amino acid histidine were mutagenized and mutants incapable of histidine biosynthesis were isolated. [Note: the original, normal strain that can grow in the absence of added histidine is said to be prototrophic for histidine, while the mutants are auxotrophs.] The mutations fell in four complementation groups, M1-M4. The ability of various compounds to rescue growth of the mutants when added to minimal growth medium is shown (+ indicates growth, - indicates lack of growth):
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| Propose a pathway for the biosynthesis of histidine. (Use M1 - M4 to denote the genes that are represented by these mutations.) | |||||||||||||||||||||||||||||||
| 10. |
The following table shows the ability of various mutant strains of Neurospora to grow on medium lacking thiamine (a vitamin) or on medium containing intermediates in the synthesis of thiamine. Again, + indicates growth, - indicates lack of growth.
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| (i) | Why do these data support a branched pathway for the synthesis of thiamine rather than a linear pathway? | ||||||||||||||||||||||||||
| (ii) | Propose a pathway for the biosynthesis of thiamine. (Use thi-1, thi-2, and thi-3 to denote the genes that are represented by these mutations.) | ||||||||||||||||||||||||||
| 11. |
A plant that normally produces purple-colored flowers was mutagenized and the following phenotypes of null mutations were observed:
Propose a genetic pathway for flower color in this plant. |
| 12. |
A different plant produces flowers that are part white and part orange. Phenotypes of various null mutations are indicated:
Suggest a genetic pathway for flower color in this plant. |
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| 13. |
Normal regulation of DNA synthesis in yeast depends on three genes, CLN, CLB, and SIC. Phenotypes of various null mutations are indicated:
[Note: In real life, the experiment would be complicated by the fact that these mutations would probably have lethal consequences. Don't let that bother you.] |
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| (i) | Which gene is absolutely required for DNA synthesis? | |||||||||||||||
| (ii) | Propose a genetic pathway for regulation of DNA synthesis in yeast. | |||||||||||||||
| Week 7 |
[Some of these are less than oven-fresh -- they're recycled from 1998, for lack of time. My apologies.]
Do Question #2, #3, and #9 for quiz section
Lecture 19
| 1. | In fruit fly, black body (b) and reduced bristle (rd) are recessive alleles of linked autosomal genes. In a strain that was b +/+ rd, it was noted that flies occasionally showed small patches of the recessive phenotype--mostly lone patches of black body and twin patches-- black body patches next to reduced bristle patches. The lone black patches and the twin patches occurred in ratio of 5:6. | |
| (i) | Explain how the patches arose, with diagrams of the chromosomes and chromatids as necessary (including the centromere). | |
| (ii) | Draw a map of the chromosome showing relative distances between the centromere and each of the two genes. | |
| (iii) | Closer examination of the flies revealed rare, lone patches of reduced bristle. Suggest two distinct mechanisms not involving point mutation by which these lone patches may have arisen. | |
| 2. | Molds (such as Aspergillus) grow as expanding disks, new growth being added at the outer edge of the colony. A consequence of this pattern of growth is that mitotic recombination, instead of giving patches as in Drosophila, causes sectoring as shown below. | |
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Genes for mold color (y+ = normal, y = yellow, recessive), colony morphology (r+ = normal, r = rough edges, recessive), growth density (g+ = normal, g = sparse, recesssive), and hyphal texture (m+ = normal, m = mottled, recessive) are known to be linked in one such mold. A haploid strain that shows all four normal phenotypes is mated to a haploid strain showing all four recessive phenotypes. When the resulting diploids were grown, most colonies showed the dominant phenotypes, but some sectors of recessive phenotype were seen: | |
What is the arrangement of these four genes with respect to the centromere? |
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| 3. | You heard in lecture about Curt Stern's observation of yellow and singed bristle spots after mitotic recombination in flies heterozygous for yellow and singed. A different strain of flies that were also heterozygous for these two loci were seen to have single spots that showed both recessive phenotypes (yellow and singed bristle in the same spot). Explain this observation. |
| 4. |
A diploid mold strain that was heterozygous for each of six recessive mutations was screened for recombinant sectors. The results are tabulated below:
Construct a map of the chromosome showing the gene order (including the centromere) and the relative map distances. |
Lectures 20-21
(These questions also appear on p.92 of the lecture notes.)
| 5. | A tumor the size of a marble, about 1 cubic centimeter in volume, may contain 109 cells. How many cell generations (starting from a single cell) are required to produce this tumor? How many cell divisions were involved? |
| 6. | Some uterine tumors consist of as many as 1011 cells. In women heterozygous for a particular X-linked gene, researchers have discovered that every cell of such a tumor has the same active X-linked allele. Explain this observation in terms of the Lyon hypothesis. |
| 7. | Although it is generally agreed that the path to malignancy is a multistep process, Weinberg and his colleagues were able to transform tissue culture cells in one step. Suggest an explanation for this apparent discrepancy. |
| 8. | The proto-oncogene erbB encodes the cell surface receptor for a growth factor. Binding of growth factor to the receptor signals the cell to divide. Speculate on how a mutation in the erbB proto-oncogene might lead to malignancy. |
| 9. | The protein product of Gene A, when active, promotes cell proliferation. In quiescent (non-proliferating) cells, the action of this gene is blocked by the action of a second gene (B). The wild type alleles of Genes A and B are a+ and b+, respectively. | |
| (a) | If a mutant allele of Gene A (allele a*) allows inappropriate entry into the cell cycle, do you expect that mutation to be dominant, gain-of-function, or recessive loss of function? | |
| (b) | If a mutant allele of Gene B (allele b*) allows inappropriate entry into the cell cycle, do you expect that mutation to be dominant, gain-of-function, or recessive loss of function? | |
| (c) | If allele a+ mutates to allele a* at a frequency of 10-5, what is the probability that a cell of cell of genotype a+a+ b+b+ starting on the pathway to cell proliferation by mutation of gene A? | |
Some tougher questions (try to do these without peeking at the answers!):
1998 Problem set 5, Q. 1, 2, 3
| Week 6 |
Do #4, #5, and #8 for Quiz section
Lecture 16
| 1. |
Shown below is the arrangement of genes on homologs in an inversion heterozygote. The lower chromosome contains the inversion. Upper case = dominant; the centromere is indicated by the circle near the right hand end. One of the progeny was found to have the following arrangement of genes on one homolog: Diagram what you think must have happened in meiosis in the parent to account for this arrangement of genes in the offspring. |
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| 2. | A true-breeding variety of plant has the phenotype MNPQRTW, while a different true-breeding variety has the phenotype mnpqrtw. The F1 progeny from a cross between the two varieties have the phenotype MNPQRTW. An F1 x F1 cross gives F2 progeny of two phenotype classes: MNPQRTW and mnpqrtw. | |
| (a) | Why is this result unexpected? | |
| (b) | Meiosis in the F1 plant was examined cytologically. In many Anaphase I cells, chromosome material was found to be stretched from one pole to the other, eventually breaking by the end of Anaphase. Suggest an explanation for this observation. Diagram the chromosomes involved. | |
| 3. |
Consider the following inversion heterozygote: Which of the following double crossover (DCO) scenarios would cause the greatest reduction in fertility, and why?
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| 4. | In Drosophila, scute bristles (sc) is recessive; the gene sc is located close to the tip of the X chromosome. Females with scute bristles (XscXsc) were mated with a normal-bristle (X+Y) male who had been X-irradiated. Unexpectedly, one of the male progeny had normal bristles (sc+). This male was mated with XscXsc females; of the 300 or so progeny, about half were scute bristle females and the other half were normal bristle males. Explain these results. |
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| (b) | Suggest a FISH experiment that you would do to distinguish between these hypotheses, including controls. | |||
| (c) | Which test would you feel more confident about, and why? | |||
Lectures 17 & 18
| 6. | (a) | Diagram all the kinds of nondisjunction of sex chromosomes that can occur in XX females and XY males in meiosis I alone, meiosis II alone, and both meiosis I and meiosis II. Assume that no more than one pair of centromeres fails to disjoin in any one step of meiosis, and show both steps of meiosis. (No need to draw the chromosomes; just use symbols such as "XX" , "XY", "XO", etc. |
| (b) | Which of the nondisjunction meioses you have listed in (a) would give an XYY zygote from normal (XX and XY) parents? (Assume that nondisjunction happens in only one of the parents.) |
| 7. | A male calico kitten was found in each of two litters of kittens. In one of the litters, the mom's fur was black and the dad's was orange; in the other litter, the mom was calico and the dad was orange. Remember that fur color is X-linked, and that orange (XR) is dominant to black (Xr). | |
| (a) | Why is the occurrence of male calico cats unusual? What kind of aberrant events could give rise to male calico cats? | |
| (b) | For each of the litters, can you tell where the aberrant event occurred, in the mom or the dad? | |
| 8. | (a) | If a mutation occurs during gametogenesis such that the X inactivation center (XIC) is deleted from an X chromosome -- would the consequences be worse for a male offspring resulting from that gamete, or a for a female offspring? Why? |
| (b) | If an XX zygote inherits one normal X chromosome and one X chromosome from which the Xist gene has been deleted, which of the two X chromosomes do you think will get inactivated as the embryo develops? Why? | |
| (c) | A current hypothesis is that the protein product of an autosomal gene protects one X chromosome from inactivation, so that one X is always active. Suppose a mutation occurs in the promoter of this hypothetical gene, such that the gene produces twice as much protein as normal. What phenotype would you expect for this strain with respect to X chromosome inactivation? | |
| (d) | Would you expect the mutant phenotype in (c) to be dominant or recessive? Why? |
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Selections from 1998 1-1998 Which of the following women would you consider to be more at risk of having a Down syndrome baby, and why?
2-1998 Give complete genetic explanations for each of these situations:
3-1998 A partial map of some genes on one arm of the Drosophila X chromosome is shown:
A new graduate student treats normal males with X-rays and mates them to fully homozygous recessive (abdefg/abdefg) females. From the progeny, he picks out the females and mates them to fully recessive (abdefg/Y) males. The resulting F2 progeny phenotypes were:
4-1998 A tall, tetraploid pea plant (genotype TTtt) is crossed to a short, tetraploid plant (tttt). Assuming that tall (T) is dominant, and that it only takes one dominant allele to give a dominant phenotype, what ratio of progeny phenotypes do you expect for this cross? (This one may take some thought -- sketching out the various outcomes of meiosis may help.)
5-1998 In pea, the genes for plant height (T = tall, t = short, recessive) and plant color (D = dark, d = light, recessive) are known to be on separate chromosomes. However, a plant breeder notices that when she crosses a particular TtDd plant with a recessive ttdd plant, the progeny consisted solely of TD and td phenotype plants in equal proportions; furthermore, there were only about half as many seeds as usual. (i) What was unusual about this result? (ii) Suggest an explanation for the results.
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Do #2, #4, and #7 for Quiz section
Lecture 13, cont'd
| 1. | Evidence from the crime scene in a criminal case is compared with a suspect's DNA with respect to five polymorphic sites (PS1 through PS5), and a perfect match is observed for both alleles of each polymorpic site. The frequencies of those alleles in the population are: PS1, 0.01 and 0.02; PS2, 0.003 and 0.01; PS3, 0.07 and 0.04; PS4, 0.13 and 0.08; PS5, 0.04 and 0.05. A perfect match (to a different suspect) is found in a separate case also; there, polymorphic sites PS6 through PS10 were tested, and the allele frequencies were: PS6, 0.2 and 0.4; PS7, 0.15 and 0.35; PS8, 0.4 and 0.3; PS9, 0.3 and 0.6; and PS10, 0.2 and 0.3. If you were serving on the jury for each case, who would feel more confident about finding guilty -- the suspect in case 1 or the suspect in case 2? Why? |
| 2. |
Shown below is the inheritance of an autosomal dominant trait. The numbers in brackets below each individual represent the repeat number at alleles of two different polymorphic loci -- PS1 (red) is shown in the first line and PS2 is shown in the second line.
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| (a) | Which of the two polymorphic sites, PS1 or PS2, shows evidence of linkage to the trait? Explain how you came to your decision. | |
| (b) | Which individual(s) show(s) evidence of recombination between that polymorphic site and the linked trait? | |
Lecture 14
| 3. | Assume that triplets beyond the location of a frameshift mutation are random with respect to the genetic code. What is the average length of a peptide (in aminoacids) you would expect a ribosome to synthesize beyond a frameshift mutation before it encounters a stop codon? |
| 4. |
Are food color additives mutagenic? To address this question, crosses very similar to those performed by Muller to detect mutations are set up in Drosophila -- i.e., starting with Xw/Xw females and X+/Y males and looking for absence of wildtype males in the F2 after mating individual F1 females with F1 males. As before,the Xw allele is on a balancer chromosome to prevent crossovers. The difference in this set of experiments is that the parental males are fed with one of four substances. One group of males is fed sugar water as a control for the rate of spontaneous mutation. Parental males in Groups 2, 3, and 4 are each fed a different food coloring agent. The results (number of crosses with red-eyed male F2 progeny and crosses failing to produce red-eyed male F2 progeny) are shown:
Which food coloring, if any, causes an increased level of mutagenesis compared to the background rate of spontaneous mutagenesis? Show your calculations. |
| 5. |  Different molecules absorb light at different wavelengths. For
any given molecule, the absorption spectrum shows the efficiency of absorption of light at different wavelengths
across the spectrum. Shown here are the absorption spectra of
DNA (solid red line) and protein (broken blue line) for different
wavelengths of UV light (nm = nanometer). If we were to plot the
efficiency of mutagenesis for these same wavelengths, what do
you think that curve woild look like? |
Lecture 15
| 6. |
|
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| 7. |
In Drosophila, a, b, c, d, e, and f are autosomal loci that are linked, but not necessarily in that order. Homozygous recessive (abcdef)/(abcdef) females are mated with a homozygous dominant (++++++/++++++) male who had been irradiated with X-rays. Most of the progeny showed the dominant (wildtype) phenotypes for all six genes. However, some of the progeny showed some recessive phenotypes as shown:
|
| Modified from 1998 |
|
|
| (i) | Assuming that all sectors resulted from point mutations, how would you explain the difference in sector size? In particular, how would you explain the half-sectored (i.e., half red, half white) colonies? |
| (ii) |
(This one is a challenge.) Counting the number of red sectors gives you an idea of how many times mutation must have occurred in the population you are sampling--but it doesn't tell you the mutation rate for the Ade locus (i.e., what percent of divisions result in mutation of the Ade locus). How might you deduce the mutation rate in the Ade locus using this sectoring assay? |
Do #4, #6, and #7 for quiz section
Lecture 11, cont'd
| 1. |
Achondroplasia is an autosomal dominant trait in humans. An unaffected boy whose parents were both affected had the following DNA constitution at a polymorphic site on chromosome 4:
-------------->>>>>>>--------------
|
The next four questions are mostly for review; some of this material was covered in the molecular biology review session on Oct. 11. If you have trouble with these questions, please talk to one of us. We can either answer specific questions or suggest some remedial reading.
| 2. | Two DNA samples (Sample A and Sample B, each containing one pure DNA species) were each cut with the restriction enzyme PstI. After gel electrophoresis of the the two cut samples, it was seen that the Sample A digest gave two DNA bands in the gel, while the digest of Sample B gave only a single band. | |
| (a) | Which DNA was circular and which was linear before the digestion? Assume that each DNA has at least one cut site for PstI. | |
| (b) | What can you conclude about circularity/linearity of the DNA substrates if you don't know whether they have any cut sites for PstI? | |
| 3. |
The diagram shows a 20 kb (kb = kilobase pair = 1000 base pairs) piece of DNA (horizontal line) with the locations of cut sites for the restriction enzyme EcoRI ("RI", marked above the horizontal line) and HindIII ("H3", below the horizontal line). The dotted vertical lines show the scale in kb.
|
|
| (a) |
In the appropriate lanes (marked i, ii, iii) of the gel outline shown, mark the locations of DNA bands you would see by gel electrophoresis if you cut (or "digest") the DNA with:
|
|
| (b) | In your gel diagram, circle those bands that will hybridize to the probe shown in the figure. | |
| 4. |
A certain bacteriophage has a genome consisting of one linear double-stranded DNA molecule. The DNA is digested with three different restriction enzymes (Ava I, Bam HI, and Cla I) in various combinations. The following DNA fragment sizes were observed by gel electrophoresis after the enzyme digests:
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Lecture 12
| 5. |
Suppose you want to PCR-amplify the gray-boxed segment of the following double-stranded DNA sequence:
|
|
| (a) | Write the sequences of the primers you would use to amplify as small a segment as possible that includes the complete grayed portion. Assume that the primers have to be 10 bases long. Specifiy the 5' and 3' ends of the primers. | |
| (b) | If you do 30 rounds of PCR, what will be the length of the majority of the products in base-pairs? (Asume that you are using the primers you have specified in (a)). | |
| 6. |
A certain autosomal recessive disorder in humans is associated with restriction fragment length polymorphisms for the enzyme Xba I as shown below:
The outer two Xba sites are present in all alleles (normal and disease) in the population, while one or both of the sites marked with asterisks are missing in the allele associated with the disorder. |
|
| (a) | How would you determine, by restriction enzyme digests and Southern blotting, if an unaffected person is a carrier for the trait? Your answer should state your expected results for a carrier vs. a homozygous unaffected person. Assume that you have an appropriate source of DNA. | |
| (b) | How many different RFLP variant types (alleles) would be expected in the population for this locus? | |
| (c) | How many different genotypes would be expected? | |
| 7. | A man who is heterozygous for an autosomal recessive trait (genotype = D/d) has 8 and 18 repeats of a microsatellite sequence at a polymorphic site (i.e., one homolog has 8 repeats and the other homolog has 18 repeats). He weds an affected woman who has 7 and 15 repeats at that same polymorphic site. | |
| (a) | Would you consider the polymorphic alleles to show a dominant/recessive behavior, an incompletely dominant behavior, or co-dominance? Why? | |
| (b) | If they have a child, what are the possible genotypes of the child, assuming that the trait is unlinked to the polymorphic locus? Include the probability of each genotype. | |
| (c) | What are the possible genotypes of the child if the trait is linked to the polymorphic site at a map distance of 20 cM? | |
| 8. | The pedigree shows an autosomal recessive trait; below the pedigree is a representation of a gel showing alleles of a polymorphic locus that is very tightly linked to the trait for the couple as well as their 8 children. Assume that no crossovers are detected in this family and fill in the phenotypes of the eight children (II-1 through II-8) with respect to the disease. Also assume that I-1 is heterozygous for the autosomal recessive trait. |
 |
Lecture 13 (to be continued in Week 5 Practice)
From 1998
| 9. | The plot below shows linkage analysis of two human disease genes with respect to a three polymorphic loci. Each curve represents the distribution of lod score vs. recombination frequency for one of the polymorphic sites with respect to one of the disease genes. Construct a linkage map that best fits the data shown. (Note: Pairs of data sets -- e.g., 'a' and 'e' -- are shown here as overlapping perfectly. You wouldn't normally expect to find this kind of perfect overlap.) |
 |
Do #2, #5, & #8 for quiz section
Lecture 9
1.
Assume that 8% of all meioses in the Great Pacific Northwestern
skunk have a crossover between the white-stripe and skunk-breath
loci. What is the map distance between these two loci?
Consider the results of the following crosses and the resulting progeny (upper case = dominant, lower case = recessive; progeny phenotypes are listed). Which loci can you assign to the same linkage group(s),
and what are the relevant map distances? What is the arrangement
of alleles (cis vs. trans) in each heterozygous parent? AaBb x aabb AaDd x aadd AaFf x aaff BbEe x bbee DdFf x ddff
2.
There are two rival factions amongst mango geneticists. One faction
holds that the taste locus (T=tart, dominant; t = sweet, recessive) is unlinked to the texture locus (F = fibrous, dominant; f = smooth, recessive). Their bitter rivals contend that the two
loci are linked at a map distance of 44 cM. To end the mango wars
once and for all, a UN Special Science Envoy steps in and does
the following experiment. A fully homozygous dominant strain is
crossed with fully recessive strain. The resulting doubly heterozygous
F1 are test -crossed, and the progeny phenotypes are: Do you think the envoy has successfully settled the question of
which faction is right? If so, which one? Provide statistical
evidence to support your answer. What would you do (to settle
the question) if you were the UN Special Science Envoy?
3.
Lecture 10
4.
Igor Young, an eager young graduate student, is working on itchy, sneezy, and jumpy, three recessive, linked traits in mice. He wants to do a three-point
test cross and construct a map of the region containing these
three genes, so he sets up the following crosses: homozygous itchy mice are crossed to mice that are itchy, sneezy and jumpy. The F1 progeny are test-crossed with itchy, jumpy, sneezy mice; he plans to count the resulting progeny and figure out
the gene order and map distances. Where is the flaw in his plan?
(Explain.) What genotypes should he be using in his final cross?
If you are told that jumpy is the middle gene, and that it is separated from itchy by 18 cM and from sneezy by 12 cM, what are the progeny you'd expect from your cross,
and how many of each would you expect, if you counted 1000 total
progeny?
Trihybrid eggplants with hairy leaves, purple flowers, and thorny
stalks are testcrossed with plants having hairless leaves, white
flowers, and smooth stalks. The progeny are: Derive a genetic map of the three loci. What is the coefficient
of coincidence?
5.
Phenotype
Number
hairy, purple, thorny
10
hairy, purple, smooth
132
hairy, white, thorny
81
hairy, white, smooth
998
hairless, purple, thorny
1020
hairless, purple, smooth
101
hairless, white, thorny
150
hairless, white, smooth
8
6.
Suppose you know of three linked loci on the Drosophila X chromosome
(say, a, b, and c). If you wanted to do a 3-point test cross to map the three loci,
how would you set up the cross? (i.e., what genotypes do you use
for males and females in the testcross?) Assume that you are starting
out with pure-breeding stocks that show all three recessive traits
and pure breeding stocks that show all three dominant traits,
and show how you derive the genotypes that you are going to use
for the testcross. What phenotype classes do you expect to see
in your testcross progeny? (Don't worry about progeny numbers;
just give the progeny classes.)
Lecture 11
Five mouse-human hybrid cell lines were examined for the presence
of human enzyme Q (which is not made by mouse cells). In the table,
"+" indicates the presence of enzyme Q activity, and "-" indicates
absence of the enzyme. Which human chromosome carries the gene
for enzyme Q?
7.
Cell line
Enzyme Q activity
Human chromosomes present in the cell line
A
-
3, 4, 7, 12
B
+
2, 6, 7, 8, 12, 14, 15
C
+
2, 6, 8, 14, 15, 18, 22
D
-
6, 14, 15
E
+
8, 9, 13, 17
The table below shows which of three human enzymes (G, AD, and
H) were present in each of five mouse-human hybrid cell lines.
Which human chromosome carries which enzyme gene?
8.
Cell line
Enzyme present
Human chromosomes present in the cell line
A
G, AD, H
1, 2, 3, 5, 9, 12, 14, 21
B
G, H
2, 3, 9, 11, 12
C
none
1, 2, 3, 12, 20
D
AD
5, 14, 20
E
G, H
2, 5, 9, 10, 15
Selections from yesteryear 1998-1 The dominant allele O is required for pigment deposition in the iris of the human eye,
while its recessive allele o causes ocular albinism. The dominant allele D is required for color perception, while its recessive allele
d is associated with color blindness. Both genes are located on
the X chromosome. (a) Assuming no crossing over, what would you predict should be
the results of a cross between a woman with ocular albinism who
is homozygous normal for color vision, and a man normal pigmentation
of the iris but who is colorblind because of the recessive d allele? (b) Assuming no crossing over (and assuming no aberrant events),
list all possible results of a cross between a woman who is heterozygous for both traits
and a man who is normal with respect to both traits.
1998-2 In Drosophila, sable body (s), singled bristle (sn), and fused vein (fu) are recessive alleles of three linked genes (wild type, dominant
alleles being s+, sn+, and fu+). Trihybrid females were crossed to males showing the three dominant
traits, and the progeny phenotypes were: (a) Give the genotypes of the parental females and male, showing
the correct sequence of genes. (b) Construct a genetic map of the region. (c) Calculate the coefficient of coincidence, and interference
(if any).
1998-3 Six mouse-human hybrid cell lines were examined by Southern blotting
for the presence or absence of human insulin gene DNA; the data
are shown below. Identify the chromosome that carries the insulin
gene.
1997-4 The following pedigree shows inheritance of red-green colorblindness
(G = normal, g = recessive, colorblind) and hemophilia (H = normal,
h = recessive, hemophilic), two X-linked traits. (a) What is the genotype of each person in the pedigree? Be as
specific as possible, giving alternative genotypes if necessary. (b) Which person is definitely a recombinant? (c) For individual III-3, what are the probabilities of being
H/H? H/h? h/h? [Note that these three probabilities must add up
to 1.] Assume that colorblindness and hemophilia are 3 map units
apart, and that III-3 is homozygous g/g.
Females
s+ sn+ fu+
1029
Males:
s+ sn+ fu+
69
s+ sn+ fu
321
s+ sn fu+
17
s sn+ fu+
99
s sn fu+
307
s sn+ fu
21
s+ sn fu
91
s sn fu
75
Note: The cell line names have been simplified, but the data are
real: they come from a real experiment, published in Nature in 1980, that determined which human chromosome has the insulin
gene.
Cell line
Human insulin sequence present?
Human chromosomes that are present in the cell line
A
Yes
6
7
10
11
14
17
18
20
21
X
B
Yes
3
5
11
14
15
17
18
21
C
Yes
4
5
10
11
12
17
18
21
D
No
8
10
12
15
17
21
X
E
No
2
5
6
10
12
18
20
21
X
F
No
17
18
20
Do #7, #8, #10, & #15 for quiz section
Lecture 5
1.
In a diploid organism where the haploid chromosome number = 9,
how many chromatids are present at:
(a)
mitotic metaphase I?
(b)
meiotic metaphase I?
(c)
meiotic metaphase II?
2.
Some organisms exhibit alternation of generations--they can exist
as haploid or as diploid forms. However, only one of these forms
undergoes meiosis. Which form is it, and why can meiosis not occur in the other form?
3.
Shown here is a pair of metaphase homologs:
(a)
If, after one division, you find that the two daughters have the
following chromosomes, was the division mitosis or meiosis?
(b)
If, after one division, you find that the two daughters have the
following chromosomes, was the division mitosis or meiosis?
Lecture 6
| 4. | Galactosemia (inability to digest milk sugar) and albinism (absence of melanin) are both inherited as recessive disorders in humans. The genes involved map to chromosomes 9 and 11, respectively. | |
| (a) | Using G/g for normal/galactosemic, A/a for normal/abino, and X/Y for X chromosome/Y chromosome, list all possible genotypes of sperm that a man who is a carrier for both disorders will produce, including the sex chromosome in your list of genotypes. | |
| (b) | The man marries a galactosemic albino woman. What is the genotype of a sperm that will result in their first son being galactosemic and albino? | |
| (c) | Diagram the meiotic divisions that produced this sperm. | |
| 5. | Hemophilia is inherited in X-linked recessive fashion; the normal allele (H) is dominant to affected (h). | |
| (a) | What are parental genotypes in the six possible matings with respect to this gene? | |
| (b) | In which of these matings are all daughters carriers? | |
| (c) | A couple has an affected daugher and an unaffected son. What are the parental genotypes? | |
| 6. | With what modes of inheritance is the disease (filled symbols) in this pedigree not consistent? What mode of inheritance do you think is probable? |  |
| 7. | With what modes of inheritance is the disease (filled symbols) in this pedigree not consistent? What mode of inheritance do you think is probable? Can you say anything about whether the disease is rare or not? |  |
Lecture 7
| 8. |
|
9.
A/a, B/b, D/d, and E/e are loci on four separate automosomes.
If a cross AABbDdee x AaBbddEe is performed, what fraction of
the progeny will:
(a)
have the phenotype ABde (where upper case = dominant, lower case
= recessive)?
(b)
have the genotype AabbddEe?
| 10. | Chickens showing the creeper phenotype have shorter than normal wing bones and leg bones. An independently segregating gene determines white vs. yellow skin color. Chickens that were fully heterozygous were mated to each other. The progeny consisted of creeper white, creeper yellow, normal white, and normal yellow chickens in 6:2:3:1 ratio. Explain the results. |
Lecture 8
| 11. | Consider a family with three children. If you know that at least 2 of the children are boys, what is the probability that all three are boys? |
| 12. | Wilson disease is a recessive disorder of copper metabolism in humans. A couple, both heterozygotes, plan to have six children. What is the probability that exactly two of the children will be affected? What is the probability that at least two of the children will be affected? |
| 13. | Purple flower color (P) in pea is dominant to white (p), and tall (T) is dominant to short (t). Plants that are heterozygous for both traits are selfed, and 3200 progeny are obtained. | |
| (a) | What progeny phenotypes do you expect to see, and how many of each phenotype do you expect? | |
| (b) |
The actual results were:
Test the goodness of fit for these results using chi-square analysis. |
|
| 14. |
Returning to purple pea flowers (dominant) vs white pea flowers (recessive) -- A seed merchant has just collected a large batch of pea seeds from what she thinks is a homozygous purple x homozygous purple cross. However, her apprentice unsettles her by claiming that the seeds really came from a heterozygote x heterozygote cross. The merchant thinks the apprentice is wrong, but decides that the best thing to do is to sample some of the seeds (by growing them up and checking the flower color). What is the minimum number of seeds she must sample to be 98% sure that the cross was not heterozygote x heterozygote? |
| 15. | This pedigree shows inheritance of an autosomal recessive trait. Assuming that individual II-4 is homozygous normal, what
is the probability that the child IV-1 will be affected?  |
Selections from 1998 1998-1 Which of the following modes of inheritance could explain this
pedigree? Give reasons. Assume that the disease is rare.
1998-2 Which of the following modes of inheritance could explain this
pedigree? Give reasons. Assume that the disease is rare.
1998-3 You were told in lecture about the primary exceptions that Calvin
Bridges noted in his cross of white-eyed female and red-eyed male
fruit flies. In a continuation of the experiment, he crossed the
white-eyed, primary exceptional females with red-eyed males. Among
the progeny were secondary exceptions--white-eyed females and
fertile, red-eyed males. Drawing on your knowledge of meiosis,
explain how these secondary exceptions might have arisen.
1998-4 A couple, both heterozygous for albinism, have five children. (a) What is the probability that the children will have the following
phenotypes in the order stated? 1st child = normal; 2nd child
= normal; 3rd child = albino; 4th child = albino; 5th child =
albino (b) What is the probability that of the five children, two will
be normal and three albino, in any order? (c) What is the probability that all five will be normal? (d) What is the probability that at least one child will be albino?
(i) autosomal recessive
(ii) autosomal dominant
(iii) X-linked recessive
(iv) X-linked dominant
(v) Y-linked
(vi) sex-influenced
(vii) sex-limited
(i) autosomal recessive
(ii) autosomal dominant
(iii) X-linked recessive
(iv) X-linked dominant
(v) Y-linked
(vi) sex-influenced
(vii) sex-limited
Do #3 and #5 for quiz section
Lectures 1-2
1.
Tall plant height (T) in the pea is dominant over short plant height (t).
(a)
If true-breeding tall plants are crossed to pure-breeding short
plants, what genotypes and phenotypes would you expect to see
in the F1?
(b)
If the F1 plants are selfed, and 1000 F2 plants are obtained,
how many of these F2 would you expect to be true-breeding when
selfed?
(c)
A tall plant of unknown history is test-crossed, and the resulting
progeny are all tall. Draw the cross, including the most probable genotypes of the
parents and offspring.
(d)
If the tall plant is known to be heterozygous, what phenotype
ratio do you expect in the progeny?
2.
Pea plants of known phenotype but unknown genotype were crossed
as shown below. What is the most probable genotype of the parents
in each cross?
Parents
Progeny
(a) tall x tall
160 tall, 0 short
(b) tall x tall
215 tall, 70 short
(c) tall x short
122 tall, 128 short
(d) tall x short
232 tall, 0 short
(e) short x short
0 tall, 171 short
3.
The tall progeny from crosses (a), (b), (c), and (d) above were selfed. In each
experiment (a-d), what fraction of the self-crosses do you expect
will yield short plants?
4.
In humans, having free-hanging earlobes is dominant over having
attached earlobes. A married couple both have free earlobes, although
they each have a parent with attached earlobes, and their only
child has attached earlobes. Draw a pedigree for this family,
indicating the genotypes where possible.
5.
Continuing with free-hanging vs. attached earlobes in humans...
(a)
List all the possible matings (parental genotypes) that could
result in a heterozygous child
(b)
Of these possible matings, which one gives the greatest proportion
of heterozygous offspring?
(c)
Which of the matings in (a) will produce offspring of only one
genotype?
(d)
Which of the matings in (a) will produce offspring of two genotypes?
(e)
Which mating is not included in either of these two lists (c and
d)?
Lectures 3-4
6.
Curly-winged Drosophila, when mated to true-breeding wildtype
(normal winged) strains, produce curly-winged and normal-winged
progeny in 1:1 ratio.
(a)
Indicate the genotypes of the two strains (make up your own allele
designations). Which phenotype is dominant?
(b)
Assume that the curly-wing condition is lethal when homozygous.
If the curly-winged F1 flies are crossed to each other (i.e.,
curly-wing x curly-wing), what fraction of the progeny will be
true-breeding?
7.
A certain DNA molecule has a base ratio of (A+G)/(C+T) = 0.81.
Is this molecule single-stranded or double-stranded? Explain.
8.
A different DNA molecule has a base ratio of (A+G)/(C+T) = 1.0.
Testing of the base composition indicates that the cytosine content
is 19%. What is the thymine content of the molecule?
9.
Consider a plant of genotype Tt. How many T alleles and t alleles
will a cell of this plant have at metaphase of mitosis?
10.
The gene coding for the beta chain of hemoglobin can have many
different alleles, the most common of which are HbA, HbS, HbE, and HbC.What are all the genotypes possible with these four alleles?
How many of them can be homozygous?
Selections from 1998 While on a hike near Mt. Rainier, you discover a patch of plants
that, according to your botanist friend, all belong to the same
species. However, the plants came in three varieties--some produce
red flowers, some blue, and some white. You surreptitiously dig
up two red-flowered plants, one blue-flowered, and one white-flowered
plant and bring them back to your lab. The results of various
crosses are as follows: Come up with a unifying hypothesis to explain these results. Your
answer must show the genotypes of all the plants concerned. Based
on your hypothesis, predict the results of cross (h).
With respect to ABO blood types, what is the one "cross" that
can give all four possible phenotypes among the progeny?
Cross
Progeny
(a) Red-flowered plant #1, selfed
3/4 red-, 1/4 blue-flowered**
(b) Red-flowered plant #2, selfed
3/4 red-, 1/4 white-flowered
(c) Blue-flowered plant, selfed
3/4 blue-, 1/4 white-flowered
(d) Red plant #1 x Red plant #2
3/4 red-, 1/4 blue-flowered
(e) Red plant #1 x blue
1/2 red-, 1/2 blue-flowered
(f) Blue-flowered x white-flowered
1/2 blue-, 1/2 white-flowered
(g) White-flowered plant, selfed
All white-flowered
(h) Red plant #2 x blue
A certain growth disorder is associated with a specific chromosomal
translocation. A prevailing hypothesis is that the translocation
splits a growth factor gene (the normal gene is depicted here),
while a minority view is that the translocation breakpoint is
more than 10 kb distant from the gene.
Six different human tissue culture cell lines were established;
each line had a duplication of a segment of chromosome 3, detected
by staining and looking at the pattern of bands. The figure shows
the normal pattern of bands 1 - 7; the cell lines had duplications
of the indicated regions (e.g., Duplication 1 repeats bands 1,
2, and 3). The cell lines were all tested for the production of
two different enzymes (E and Z). Normal cells produce 60 units
of Enzyme E and 100 units of Enzyme Z. The enzyme levels in the
six duplication cell lines are shown below. Assuming that the
level of enzyme scales linearly with the number of copies of the
gene coding for the enzyme, which band most likely has the gene
for Enzyme E? For Enzyme Z?
